Let $E$ be a vector bundle over a smooth manifold $M$ equipped with a linear connection $\nabla : \Gamma(E) \to \Omega^1(M;E).$ I say $(M,E,\nabla)$ is flat if it admits trivial local models; i.e. if for each $p \in M$ there is a $\nabla$-parallel local frame for $E$ defined on some neighbourhood of $p$. It is well known (and often instead taken as the definition) that $(M,E,\nabla)$ is flat if and only if the curvature form $R^\nabla \in \Omega^2(M; E)$ vanishes; so the curvature can be motivated as an obstruction to flatness.

When $E = TM$ so that $\nabla$ is an affine connection, a more restrictive definition is often used: we say $(M,\nabla)$ is flat if each $p \in M$ is contained in a chart whose coordinate frame is $\nabla$-parallel. This imposes an additional requirement on the local model: not only must we be able to choose a frame making $\nabla$ trivial, but this frame must be holonomic. Again, a nice characterization of this kind of flatness is well-known: it's equivalent to both the curvature $R^\nabla$ and the torsion $T^\nabla$ vanishing.

Thus in the world of affine connections that are flat in the weaker sense ($R^\nabla = 0$), torsion has a very simple motivation: it is exactly the obstruction to the existence of trivial local models, by which I mean charts $x^i$ such that $\nabla \partial_i=0.$ One way to think about this is that torsion is an obstruction to the integrability of frames: thanks to $R^\nabla = 0$ we can find a parallel frame $e_i$, and $T^\nabla = 0$ (implying $[e_i,e_j] = 0$) is then exactly the condition guaranteeing that $e_i = \partial_i$ for some chart.

Question. Without the assumption that $R^\nabla = 0$, can we motivate torsion as the obstruction to some kind of integrability?

This is motivated in part by this nice answer on MO, which describes torsion as an obstruction to the integrability of various $G$-structures; but I don't see how this interpretation can apply in the case of a connection alone.

One vague notion bouncing around in my head is some kind of Poincaré lemma for the solder form $\mathrm{id} \in \Omega^1(M;TM)$: we know $d^\nabla \mathrm{id}$ is the torsion, so perhaps vanishing torsion implies the solder form is locally the covariant derivative of some vector field? I'm not sure if this is actually true (I've only seen covariant exterior calculus treated for flat connections, since this is where you get a de Rham complex), nor how best to interpret it if it is.

Cross-posted on MO.

  • $\begingroup$ The tricky thing about torsion is it pretty much affects the structure of any compatible geometric structure you're trying to define over your manifold (sometimes in ways that cannot be rectified). Maybe you've seen this already and I'm sorry for the self promotion but maybe my input on the question in MO can help. In some (very loose) sense I think torsion is the obstruction to making sense of second derivatives. $\endgroup$ Dec 28, 2017 at 21:54
  • $\begingroup$ @SaalHardali: thanks for the note - even though it's not quite what I'd hoped for, this perspective does somewhat decouple the curvature and the torsion, which is what I'm trying to get at. $\endgroup$ Dec 29, 2017 at 10:09

1 Answer 1


The torsion of an affine connection is pointwise, in the sense that if two affine connections, $\nabla$ and $\nabla',$ agree at a point $p$, then so do their respective torsions. This property distinguishes torsion from curvature, for example, as the curvature of a connection at a point depends on the connection on a neighborhood. Therefore, it makes sense that torsion is the obstruction for some pointwise property, as opposed to integrability of some sort.

Indeed, we have the following simple observation. The torsion of an affine connection $\nabla$ vanishes at a point $p$ if and only if there exist local coordinates around $p$ whose induced affine connection coincides with $\nabla$ at $p$. The "if" part is trivial, as any connection induced by coordinates is torsion-free. Conversely, suppose the torsion of $\nabla$ vanishes at $p$. Let $\exp_p$ denote the exponential map with respect to $\nabla$, and use it as a local parametrization around $p$. Let $\Gamma_{ij}^k$ denote the Christoffel symbols of $\nabla$ with respect to this parametrization. By construction, any curve $\gamma$ whose coordinate expression with respect to $\exp_p$ is a straight line through the origin is a geodesic passing through $p$. As the geodesic equation in coordinates reads $$\ddot{\gamma}^k+\dot{\gamma}^i\dot{\gamma}^j\Gamma_{ij}^k=0,$$this means that for any tangent vector $v=v^i\partial/\partial x^i\in T_pM$ and $k=1,\ldots,n$ we have $$v^iv^j\Gamma_{ij}^k=0.$$In other words, the tensor $\Gamma$ is anti-symmetric at $p$. But, by assumption, $\nabla$ is torsion-free, which means that $\Gamma$ is symmetric, and the proof is complete.

The above observation can also be thought of as the following, more geometric, interpretation. Let $M$ be a smooth manifold, let $p\in M$, and let $\nabla$ be an affine connection on $M$. Let $L\subset T_pM$ be a linear subspace, and let $U\subset L$ be a neighborhood of $0$, such that $\exp_p|_U$ is an immersion, where $\exp_p$ is taken with respect to $\nabla$. Write $\Lambda:=\exp_p(U)$. If the torsion of $\nabla$ vanishes at $p$, then $\Lambda$ is "totally geodesic" at $p$, in the sense that we have $$\nabla_XY(p)\in L=T_p\Lambda$$ for any two vector fields $X$ and $Y$ which restrict to vector fields on $\Lambda$. (This follows from the above observation by taking exponential coordinates around $p$).


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